Why is enthalpy of neutralisation nearly constant for strong acid + strong base?

Because the net reaction is always H⁺(aq) + OH⁻(aq) → H₂O(l) with ΔH ≈ −57.3 kJ/mol; the spectator ions do not contribute.

Is ΔH always equal to heat released?

Only at constant pressure with no non-PV work. Under constant volume conditions, q equals ΔU, not ΔH.

Can a reaction be endothermic and spontaneous?

Yes — if the entropy increase is large enough, ΔG = ΔH − TΔS can still be negative. NH₄NO₃ dissolving in water is a classic example.

Why are bond enthalpies called 'average'?

The same bond (e.g. C-H) has slightly different dissociation energies in different molecules; tabulated values are averages over many compounds.

What is the significance of a Born-Haber cycle?

It uses Hess's law to calculate lattice energies of ionic solids that cannot be measured directly, by combining sublimation, ionisation, electron affinity, and bond-dissociation steps.

ΔH_f° of O₂(g) at 298 K is:

0 kJ/mol. By convention the standard enthalpy of formation of an element in its standard state is zero. O₂ at 1 bar, 298 K is the standard state of oxygen.

For C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ. The enthalpy released when 12 g of C burns is:

−393.5 kJ. 12 g of C is exactly 1 mol, releasing 393.5 kJ.

Which is true at constant pressure?

q = ΔH. At constant P, q_p = ΔH because ΔH = ΔU + PΔV = ΔU + w_pV.

Using bond enthalpies (H-H = 436, Cl-Cl = 243, H-Cl = 432 kJ/mol), ΔH for H₂ + Cl₂ → 2 HCl is:

−185 kJ. Bonds broken (679) − bonds formed (864) = −185 kJ.

A reaction has ΔH = −100 kJ/mol and Δn_gas = +1 at 300 K. ΔU is approximately:

−102.5 kJ. ΔU = ΔH − Δn_g RT = −100 − (1)(8.314×300/1000) = −102.5 kJ/mol.

Chemical Energetics MDCAT 2026 — Notes, Key Concepts & MCQs (Chemistry)

Chemical Energetics

Chemical Energetics averages 2 MCQs per MDCAT paper — Hess's law, enthalpies of formation/combustion, and bond enthalpies are most tested.

Chemical Energetics is a Chemistry chapter on the official PMDC MDCAT 2026 syllabus, contributing roughly 2 MCQs to the 45-MCQ Chemistry section. Mastering the core concepts below typically secures the full chapter weightage.

System, surroundings, and the first law

A thermodynamic system is the part of the universe under study; the rest is the surroundings. The first law: ΔU = q + w, where q is heat added to the system and w is work done on it. At constant volume, q_v = ΔU; at constant pressure, q_p = ΔH, where H = U + PV. Most laboratory reactions occur at constant atmospheric pressure, so chemists tabulate ΔH. Atkins Chapter 2 derives ΔH = ΔU + Δn_gRT for gas-phase reactions — vital when converting between calorimetry data and tabulated values.

Enthalpy changes — formation, combustion, neutralisation

ΔH_f° is the enthalpy of formation of one mole of compound from its elements in standard states (1 bar, usually 298 K). By convention, ΔH_f°(element in standard state) = 0; this is why ΔH_f°(O₂) = 0 but ΔH_f°(O₃) = +142.7 kJ/mol. Combustion enthalpies are always negative (exothermic): CH₄ + 2 O₂ → CO₂ + 2 H₂O, ΔH = −890 kJ/mol. Strong-acid–strong-base neutralisation gives a near-constant −57.3 kJ/mol because the only chemistry is H⁺(aq) + OH⁻(aq) → H₂O(l).

Hess's law and enthalpy cycles

Because H is a state function, the total enthalpy change of a multi-step path equals that of any other path between the same endpoints. ΔH_rxn° = ΣΔH_f°(products) − ΣΔH_f°(reactants). For 2 H₂(g) + O₂(g) → 2 H₂O(l), ΔH = 2(−285.8) − 0 − 0 = −571.6 kJ. The Born-Haber cycle applies the same idea to ionic-lattice formation: atomisation, ionisation, electron affinity, lattice energy. NaCl: lattice energy ≈ −787 kJ/mol — too large to measure directly, so Hess's law lets us calculate it.

Bond enthalpies and reaction energetics

Average bond enthalpy is the energy per mole to break one mole of a particular bond in the gas phase. ΔH_rxn ≈ Σ(bonds broken) − Σ(bonds formed). For H₂ + Cl₂ → 2 HCl: bonds broken = 436 + 243 = 679 kJ; bonds formed = 2 × 432 = 864 kJ; ΔH = 679 − 864 = −185 kJ. The approximation works because C-H, O-H bond energies vary by only a few kJ across compounds — but it is approximate, not exact. Clayden uses bond-dissociation enthalpies extensively to predict reaction feasibility.

Calorimetry and a worked numerical

In bomb calorimetry (constant volume), q_v = C_cal ΔT gives ΔU directly; coffee-cup calorimetry at constant pressure gives ΔH. If 0.500 g of glucose (M = 180) burned in a calorimeter (C_cal = 5.00 kJ/K) raises the temperature by 1.56 K, q = 7.80 kJ for 2.78×10⁻³ mol, so ΔU_combustion = −2806 kJ/mol — close to the literature −2803 kJ/mol. The MDCAT often gives mass and ΔT and asks you to back out ΔH per mole; convert grams to moles first.

Key Concepts

Worked MCQs

Frequently Asked Questions

How Chemical Energetics Is Tested

MDCAT questions on Chemical Energetics are a mix of recall (definitions, classifications), application (predict outcomes, interpret diagrams), and basic numerical/analytical reasoning. PMDC papers from 2020–2025 emphasized the concepts above; older UHS papers (2008–2019) tested them too, with slight variations in question framing.